What is the integral of ln(x)x^{-1} ?

The integral of ln(x)x^{-1} is (ln^2(x))/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ln(x)x^{-1}

\int ln (x) x^{- 1} d x

\frac{ln ^{2} ( x )}{2} + C

Solution steps

\int ln (x) x^{- 1} d x

Simplify

ln (x) x^{- 1} : \frac{ln ( x )}{x}

= \int \frac{ln ( x )}{x} d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{u ^{2}}{2}

Substitute back

u = ln (x)

= \frac{ln ^{2} ( x )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( x )}{2} + C

\int \frac{x ^{3} - 6 x + 5}{3 x ^{2} - 3 x - 18} d x

\int (1 + sin^{2} (x)) cos (x) d x

\int (4 e^{4})^{x} d x

\int \frac{n}{n ^{2} + 1}

\int \frac{2 x ^{3}}{( x ^{2} + 1 ) ^{3}} d x

What is the integral of ln(x)x^{-1} ?
The integral of ln(x)x^{-1} is (ln^2(x))/2+C